Question

prove the following identities:

(1 - sinA - cosA)= 2(1-sinA) (1-cosA)

really need help

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Answer 1

Appropriate Question :-

Prove the following :-

$\rm \: {(1 - sinA - cosA)}^{2} = 2(1 - sinA)(1 - cosA)$

$\purple{\large\underline{\sf{Solution-}}}$

Consider, LHS

$\rm :\longmapsto\: \rm \: {(1 - sinA - cosA)}^{2}$

We know,

$\boxed{ \bf{ \: {(x + y + z)}^{2} = {x}^{2} + {y}^{2} + {z}^{2} + 2xy + 2yz + 2zx}}$

So, using this identity, we get

$\rm \: = {1}^{2} + {( - sinA)}^{2} + {( - cosA)}^{2} + 2(1)( - sinA) + 2( - sinA)( - cosA) + 2( - cosA)(1)$

$\rm \: = \:1 + {sin}^{2}A + {cos}^{2}A - 2sinA + 2sinAcosA - 2cosA$

$\rm \: = \:1 + ( {sin}^{2}A + {cos}^{2}A) - 2sinA + 2sinAcosA - 2cosA$

We know,

$\boxed{ \bf{ \: {sin}^{2}x + {cos}^{2}x = 1}}$

So, using this identity,

$\rm \: = \:1 + 1 - 2sinA + 2sinAcosA - 2cosA$

$\rm \: = \:2 - 2sinA + 2sinAcosA - 2cosA$

$\rm \: = \:2\bigg[1 - sinA + sinAcosA - cosA\bigg]$

$\rm \: = \:2\bigg[1(1 - sinA) - cosA(1 - sinA)\bigg]$

$\rm \: = \:2\bigg[(1 - sinA)(1 - cosA)\bigg]$

$\rm \: = \:2(1 - sinA)(1 - cosA)$

Hence,

$\boxed{ \bf{\: {(1 - sinA - cosA)}^{2} = 2(1 - sinA)(1 - cosA)}}$

Additional Information:-

Relationship between sides and T ratios

sin θ = Opposite Side/Hypotenuse

cos θ = Adjacent Side/Hypotenuse

tan θ = Opposite Side/Adjacent Side

sec θ = Hypotenuse/Adjacent Side

cosec θ = Hypotenuse/Opposite Side

cot θ = Adjacent Side/Opposite Side

Reciprocal Identities

cosec θ = 1/sin θ

sec θ = 1/cos θ

cot θ = 1/tan θ

sin θ = 1/cosec θ

cos θ = 1/sec θ

tan θ = 1/cot θ

Co-function Identities

sin (90°−x) = cos x

cos (90°−x) = sin x

tan (90°−x) = cot x

cot (90°−x) = tan x

sec (90°−x) = cosec x

cosec (90°−x) = sec x

Fundamental Trigonometric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

cosec²θ - cot²θ = 1